[tex] \frac{4}{100}* \frac{n(n+1)}{2}=n+1 [/tex]
[tex] \frac{4}{100}* \frac{n^2+n}{2}=n+1 [/tex]
[tex] \frac{4n^2+4n}{200}=n+1 [/tex]
simplificam membrul stang cu 4 si pe cel drept il amplificam cu 50 ptu a aduce la acelasi numitor
[tex] \frac{n^2+n}{50}= \frac{50n+50}{50} [/tex]
[tex]n^2+n=50n+50[/tex]
[tex]n^2+n-50n-50=0[/tex]
[tex](n^2+n)-(50n+50)=0[/tex]
[tex]n(n+1)-50(n+1)=0[/tex]
[tex](n-50)(n+1)=0[/tex]
n-50=0, n=50
n+1=0, n=-1
Deoarece n∈N doar n=50 e solutie a problemei.
